一些不等积分的练习（持续更新）

题目地址

最近会持续更新做一些这里面的题目，提供几个网站可以找答案：wolfram ，integral-calculator

1.求$int frac {1+x^2}{1+x^4}dx$

注意到以下事实$intfrac{1}{x^2+a^2}dx=frac{1}{a}arctan(frac{x}{a})+C$

故$int frac {1+x^2}{1+x^4}dx\=intfrac{1}{2}(frac{1}{x^2-sqrt{2}x+1}+frac{1}{x^2+sqrt{2}x+1})dx\=frac{1}{2}(intfrac{1}{(x-frac{1}{sqrt{2}})^2+frac{1}{2}}d(x-frac{1}{sqrt2})+intfrac{1}{(x+frac{1}{sqrt2})^2+frac{1}{2}}d(x+frac{1}{sqrt2}))\=frac{sqrt{2}}{2}(arctan(sqrt2x-1)+arctan(sqrt2x+1))+C$

2.求 $intfrac{dx}{(x+1)(x^2+1)}$

注意到$int frac{dx}{a^2-x^2}=frac{1}{a}arctanh(frac{x}{a})+C=frac{ln(|frac{x+a}{-x+a}|)}{2a}+C\arctanh(x)=frac{1}{2}(ln(x+1)-ln(1-x))$

原式$=intfrac{(x-1)dx}{x^4-1}\=intfrac{d(x^2)}{2(x^4-1)}-intfrac{dx}{x^4-1}\=-frac{1}{2}arctanh(x^2)-frac{1}{2}int(frac1{x^2-1}-frac1{x^2+1})dx\=-frac{1}{2}arctanh(x^2)+frac12arctanh(x)+frac12arctan(x)+C$

3.求$int frac {dx}{(1+x^2)(1+x^3)}$

原式$=intfrac{dx}{(1+x^2)(1+x)(1-x+x^2)}\=-frac{1}{3}intfrac{2x-1}{x^2-x+1}dx+frac{1}{2}int frac{x+1}{x^2+1}dx+frac16intfrac{1}{1+x}dx\=-frac{1}{3}ln|x^2-x+1|+frac16ln|1+x|+frac{1}{2}arctanx+frac{1}{4}ln|x^2+1|+C$

4.求$int frac{dx}{sqrt[3]{1-x^3}}$

令$t=frac{x}{sqrt[3]{1-x^3}}$，那么$dt=frac{dx}{(sqrt[3]{1-x^3})^4}$

原式$=int (1-x^3)dt=int frac{dt}{t^3+1}=-frac12ln|t^2-t+1|+frac{1}{sqrt3}arctanfrac{2t-1}{sqrt3}+frac{ln|t^3+1|}{3}+C$

5.求$int frac{dx}{lambda+sqrt{1-x^2}}$

当$|lambda|>1$时,令$x=sint$

原式$=int frac{lambda dx}{lambda^2-1+x^2}-int frac{sqrt{1-x^2}dx}{lambda^2+x^2-1}=frac{lambda}{sqrt{lambda^2-1}}arctan{frac{x}{sqrt{lambda^2-1}}}-intfrac{cos^2t}{lambda^2-cos^2t}dt=frac{lambda}{sqrt{lambda^2-1}}arctan{frac{x}{sqrt{lambda^2-1}}}+t-frac{arctanfrac{tant}{sqrt{lambda^2-1}}}{sqrt{lambda^2-1}}+C$

当$lambda=1$时,显然$原式=arcsinx+C$

当$|lambda|<1$时，原式$=int frac{lambda dx}{lambda^2-1+x^2}-int frac{sqrt{1-x^2}dx}{lambda^2+x^2-1}=frac{lambda}{sqrt{1-lambda^2}}arctanhfrac{x}{sqrt{1-lambda^2}}+t-frac{arctanhfrac{tant}{sqrt{1-lambda^2}}}{sqrt{1-lambda^2}}+C$

6.求$int frac{dx}{asinx+bcosx}$

原式$=frac{1}{sqrt{a^2+b^2}}intfrac{d(x+phi)}{cos(x+phi)}=frac{1}{sqrt{a^2+b^2}}ln|sec(x+phi)+tan(x+phi)|+C$

7.求$int ln(sqrt{x+1}+sqrt{1-x})dx$

令$m=sqrt{x+1},n=sqrt{-x+1}$

原式$=int ln(n+m)dx\=xln(n+m)-int xdln(n+m)$

记$I=int xdln(n+m)$

注意到 $2x=m^2-n^2$ $m^2+n^2=2$

我们有$I$ $=int frac{m^2-n2}{2}*frac{dn+dm}{m+n}\=intfrac{(dm+dn)(m-n)}{2}\=frac12int (mdm-ndn)+int frac{ n^2d(frac{n}{m})}{m^2+n^2}=frac14(m^2-n^2)+arctan(frac nm)+C$

将$n,m$带入，有$原式=xln(sqrt{x+1}+sqrt{x-1})-frac{x}{2}-arctan(frac{sqrt{x-1}}{sqrt{x+1}})+C$

8.求$int ln(sqrt{1-x}-sqrt x)dx$ // unfinished

21.求$int sqrt{tanx+2}dx$

令$n=tanx,m=sqrt{n^2+2}$

我们有 $mdm=ndn,frac{dn}{n^2+1}=dx$

原式$=int mdx\=int frac{mdn}{n^2+1}\=int frac{2mdn}{m^2+n^2}$

我们又注意到$int frac{mdn-ndm}{m^2+n^2}=arctan(frac{n}{m})+C$

​ $int frac{mdn+ndm}{n^2+m^2}=intfrac{mdn+frac{n^2}{m}dn}{m^2+n^2}=intfrac{dn}{m}$

不难发现$frac{dn}{m}=frac{dm}{n}=frac{dn+dm}{m+n}$

所以 $int frac{mdn+ndm}{n^2+m^2}=intfrac{mdn+frac{n^2}{m}dn}{m^2+n^2}=intfrac{dn}{m}=int frac{dm+dn}{n+m}=ln(n+m)+C$

故原式$=intfrac{mdn-ndm}{m^2+n^2}+intfrac{mdn+ndm}{n^2+m^2}=arctan(frac{n}{m})+ln(n+m)+C=arctan(frac{tanx}{sqrt{tan^2x+2}})+ln(tanx+sqrt{tan^2x+2})+C$